Home > Calculus > Identifying Stationary Points (Critical Points) for a Function

How do you find the critical points of #g'(x)=3x^2-6x^2#?

Answer 1

I can't tell you what the coordinate of your critical point is for that equation because I don't know what your original function is.

To find the critical point of an equation# f(x)#, you find its derivative #f'(x)# and set it equal to 0 and find the x-values. Then you plug those x-values into #f(x)# to get the coordinates of the critical values.

Remember, the critical value is at a point in the graph where there is a min/max.

I can't tell you what the coordinate of your critical point is for that equation because I don't know what your original function is.

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Answer 2

Charles Anctil

To find the critical points of ( g'(x) = 3x^2 - 6x^2 ), you first need to find the values of ( x ) where the derivative equals zero or is undefined.

( g'(x) = 3x^2 - 6x^2 ) simplifies to ( g'(x) = -3x^2 ).

Set ( g'(x) ) equal to zero and solve for ( x ): [ -3x^2 = 0 ]

This gives ( x = 0 ) as the only solution.

So, the critical point of ( g'(x) ) is ( x = 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.